Isometries of the Complex Plane: Fixed Lines

I am trying to find the fixed line in isometries of the plane, and I keep running into contradictions. I have checked both my assumptions and workings 3 times and still get contradictions. I'll start from the beginning, stating all my assumptions, because I do not know where I am going wrong:

I know there are 4 types of isometries: rotations, translations, reflections and glide reflections. (A glide reflection is a reflection, followed by a translation in a direction parallel to the line of reflection).

All isometries can be written as f(z) = az+b or aZ+b, with a,b in the set of complex numbers, lal=1. (I am using capital Z for z conjugate. If you know how to type z conjugate please tell me)

Isometries in the form f(z) = aZ+b are either reflections or glide reflections.
If f(b) = 0, f is a reflection.
Otherwise f is a glide reflection.

I have an isometry that gives a glide reflection, and I am trying to find the line of reflection. I am assuming that the line of reflection in a glide reflection isometry is the only fixed line.

To find the fixed line in a glide reflection, I am using this formula:
f(f(a))-2f(a)+a = 0
All values of a which satisfy this equation should lie on the fixed line. I am aware that this only works for glide reflections, not general functions.

The function I have is f(z)=iZ-2.
f(-2) = -2i -2 =/= 0, so f is a glide reflection, so using the formula above should give me the equation of the fixed line.

You have f(x+ yi)= x- yi- 2 and a line, in the complex plane, must be of the form y= mx+ c. Saying that a line is fixed means that (That is, that one point on the line is mapped into itself or another point on the line). In fact, it is sufficient to start by using the fact that a specific point on the line, (0, c) is simplest, is mapped into a point (x, mx+ c).

Setting the real parts equal, we have x= 2 and, setting the imaginary parts equal, so that c= -m. That is, the line is of the form y= mx- m. Now look at the next easiest point, (1, 0). f(1)= 1- 2= -1. But (-1, 0) will be on the line y= mx- m only is 0= -m-m= -2m: m= 0.

This glide reflection fixes the real axis. Actually, that should have been obvious since reflects around the real axis and then "- 2" translates parallel to the real axis.

By the way, I got by using LaTex. The code is [ math ]\overline{z}[ /math ] (without the spaces).

Secondly, where do you get the point (1,0) from?
f(1) = i - 2
f(i - 2) = -1 -2i
The points (1,0) (-2,1) and (-1,-2) are not colinear so (1,0) is not on a fixed line. I know that the fixed line is not parallel to either axis.

It would help if someone could tell me where the contradiction arises in my reasoning:

(i) my assumption that the function f(z) = i -2 is a glide reflection isometry of the complex plane

(ii) my assumption that a glide reflection always has exactly 1 fixed line, and that this line is the line of reflection prior to translation. (The line stays the same, but loses some of the properties that a line of reflection has). This is because a glide reflection is a reflection in a line, followed by a translation parallel to that line.

(iii) my assumption that solving the equation f(f(a)) - 2f(a) + a = 0 for a will find the values of a which lie on the fixed line, if f is a glide reflection.

(iv) my calcultion to find the values of a which solve the equation when f(z) = i -2. I found no solutions, meaning that the mapping of f contains no fixed lines. Contradion!